In this experiment, you will investigate total internal reflection using a semicircular glass block. You directs a narrow ray of light from a ray box into t...

(a) From Fig. 5.2, the critical angle c is measured between the incident ray and the normal at the flat surface at point O. Using a protractor on the diagram:

\[ c = 42^\circ \] [1]

(Accept 41° to 43°.)

(b) Calculating the sine of the critical angle:

\[ \sin c = \sin 42^\circ = 0.669 \] [1]

(Accept a value consistent with the measured angle.)

(c) The refractive index of the glass is found using \( n = 1 / \sin c \):

\[ n = \frac{1}{\sin c} \] [1]

\[ n = \frac{1}{0.669} = 1.49 \] [1]

(Accept 1.4 to 1.6 if consistent with the measured critical angle.) This value is typical for glass, confirming the measurement is reasonable.

(d) The ray must enter through the curved surface because at the curved surface, the ray travels along a radius of the semicircle, which means it hits the curved surface at right angles (perpendicular to the surface). A ray striking a surface perpendicularly does not refract, so it passes straight through to the flat surface without changing direction. [1]

This ensures that refraction occurs only at the flat surface, so the measured angle of incidence at the flat surface is the true angle.

(e) When the angle of incidence at the flat surface is less than the critical angle:

• The ray is partly refracted as it passes from the glass into the air. The refracted ray bends away from the normal (since glass is optically denser than air). [1]
• At the same time, a weak reflected ray is also produced inside the glass block. This partial reflection always accompanies refraction. [1]

(f) As the angle of incidence increases towards the critical angle, the refracted ray becomes dimmer (less bright). [1]

More and more of the light energy is reflected rather than refracted. At the critical angle, the refracted ray just grazes along the surface (at 90° to the normal), and beyond this angle, all the light is reflected.

(g) One practical use of total internal reflection (any one): [1]

• Optical fibres for telecommunications or medical endoscopes: light is guided along the fibre by repeated total internal reflection.
• Reflecting prisms in binoculars, periscopes, or SLR cameras.
• Bicycle reflectors and road reflectors.

(h) For diamond, the critical angle is 24°:

\[ n = \frac{1}{\sin 24^\circ} = \frac{1}{0.407} = 2.46 \] [1]

Diamond has a much higher refractive index than glass (2.46 compared to about 1.5), which means it has a much smaller critical angle. This causes extensive internal reflection, producing the characteristic sparkle of a cut diamond.